RS Aggarwal Solutions Class 10 Ex 10D

SOLVE EACH OF THE FOLLOWING EQUATIONS BY USING THE METHOD OF COMPLETING THE SQUARE:

QUESTION-1:

\(x^{2}-6x+3=0\)

Solution:

\(x^{2}-6x+3=0\)

\(\Rightarrow x^{2}-6x=-3\)

\(\Rightarrow x^{2}-2\times x\times 3+ 3^{2}=-3+3^{2}\) (Adding \(3^{2}\) on both sides)

\(\Rightarrow \left ( x-3 \right )^{2}=-3+9=6\)

\(\Rightarrow x-3 =\pm \sqrt{6}\) (Taking square root on both sides)

\(\Rightarrow x-3 = \sqrt{6}\:or\:x-3=-\sqrt{6}\)

\(\Rightarrow x=3+\sqrt{6}\:or\:x=3-\sqrt{6}\)

Hence, \(3+\sqrt{6}\:and\:3-\sqrt{6}\) are the roots of the given equation.

QUESTION-2:

\(x^{2}-4x+1=0\)

Solution:

\(x^{2}-4x+1=0\)

=> \(x^{2}-4x=-1\)

=> \(x^{2}-2\times x\times 2+2^{2}=-1+2^{2}\) (Adding \(2^{2}\) on both sides)

=> \(\left ( x-2 \right )^{2}=-1+4=3\)

=> \(\Rightarrow x-2 =\pm \sqrt{3}\) (Taking square root on both sides)

\(\Rightarrow x-2 = \sqrt{3}\:or\:x-2=-\sqrt{3}\)

\(\Rightarrow x=2+\sqrt{3}\:or\:x=2-\sqrt{3}\)

Hence, \(2+\sqrt{3}\:and\:2-\sqrt{3}\) are the roots of the given equation.

QUESTION-3:

\(x^{2}+8x-2=0\)

Solution:

=> \(x^{2}+8x=2\)

=> \(x^{2}+2\times x\times 4+4^{2}=2+4^{2}\) (Adding \(4^{2}\) on both sides)

=> \(\left ( x+4 \right )^{2}=2+16=18\)

=> \(x+4=\pm \sqrt{18}=\pm 3\sqrt{2}\) (Taking square root on both sides)

=> \(x+4= 3\sqrt{2}\:or\:x+4=-3\sqrt{2}\)

=> \(x=-4+ 3\sqrt{2}\:or\:x=-4-3\sqrt{2}\)

=> Hence, \(\left ( -4+3\sqrt{2} \right )\:and\:\left ( -4-3\sqrt{2} \right )\) are the roots of the given equation.

QUESTION-4:

\(4x^{2}+4\sqrt{3}x+3=0\)

Solution:

\(4x^{2}+4\sqrt{3}x+3=0\)

=> \(4x^{2}+4\sqrt{3}x=-3\)

\(\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times \sqrt{3}+\left ( \sqrt{3} \right )^{2}\) (Adding \(\left ( \sqrt{3} \right )^{2}\) on both sides)

\(\Rightarrow \left ( 2x+\sqrt{3} \right )^{2}=-3+3=0\)

\(\Rightarrow 2x+\sqrt{3}=0\)

\(\Rightarrow x=-\frac{\sqrt{3}}{2}\)

Hence, \(-\frac{\sqrt{3}}{2}\) is the repeated root of the given equation.

QUESTION-5:

\(2x^{2}+5x-3=0\)

Solution:

\(2x^{2}+5x-3=0\)

\(\Rightarrow 4x^{2}+10x-6=0\) (Multiplying both sides by 2)

\(\Rightarrow 4x^{2}+10x=6\)

\(\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times \frac{5}{2}+\left ( \frac{5}{2} \right )^{2}\) [Adding \(\left ( \frac{5}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left ( 2x+\frac{5}{2} \right )^{2}=6+\frac{25}{4}=\frac{24+25}{4}=\frac{49}{4}=\left ( \frac{7}{2} \right )^{2}\)

\(\Rightarrow 2x+\frac{5}{2} =\pm \frac{7}{2}\) (Taking square root on both sides)

\(\Rightarrow 2x+\frac{5}{2} = \frac{7}{2}\:or\:2x+\frac{5}{2}=-\frac{7}{2}\)

\(\Rightarrow 2x=\frac{7}{2}-\frac{5}{2}=\frac{2}{2}=1\:or\:2x=-\frac{7}{2}-\frac{5}{2}=-\frac{12}{2}=-6\)

=> x = \(\frac{1}{2}\) or x = -3

Hence, \(\frac{1}{2}\) and -3 are the roots of the given equation.

QUESTION-6:

\(3x^{2}-x-2=0\)

Solution:

\(3x^{2}-x-2=0\)

\(9x^{2}-3x-6=0\)

\(9x^{2}-3x=6\)

\(\Rightarrow \left ( 3x \right )^{2}-2\times 3x\times \frac{1}{2}+\left ( \frac{1}{2} \right )^{2}=6+\left ( \frac{1}{2} \right )^{2}\) [Adding \(\left ( \frac{1}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left ( 3x-\frac{1}{2} \right )^{2}=6+\frac{1}{4}=\frac{25}{4}=\left ( \frac{5}{2} \right )^{2}\)

\(\Rightarrow 3x-\frac{1}{2}=\pm \frac{5}{2}\) (Taking square root on both sides)

\(\Rightarrow 3x-\frac{1}{2}=\frac{5}{2}\:or\:3x-\frac{1}{2}=-\frac{5}{2}\)

\(\Rightarrow 3x=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3\:or\:3x=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=-2\)

=> x = 1 or x = \(-\frac{2}{3}\)

Hence, 1 and \(-\frac{2}{3}\) are the roots of the given equation.

QUESTION-7:

\(8x^{2}-14x-15=0\)

Solution:

\(8x^{2}-14x-15=0\)

\(\Rightarrow 16x^{2}-28x-30=0\) (Multiplying both sides by 2)

\(\Rightarrow 16x^{2}-28x=30\)

\(\Rightarrow \left ( 4x \right )^{2}-2\times 4x\times \frac{7}{2}+\left ( \frac{7}{2} \right )^{2}=30+\left ( \frac{7}{2} \right )^{2}\) [Adding \(\left ( \frac{7}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left ( 4x-\frac{7}{2} \right )^{2}=30+\frac{49}{4}=\frac{169}{4}=\left ( \frac{13}{2} \right )^{2}\)

\(\Rightarrow 4x-\frac{7}{2}=\pm \frac{13}{2}\) (Taking square root on both sides)

\(\Rightarrow 4x-\frac{7}{2}=\frac{13}{2}\:or\:4x-\frac{7}{2}=-\frac{13}{2}\)

\(\Rightarrow 4x=\frac{13}{2}+\frac{7}{2}=\frac{20}{2}=10\:or\:4x=-\frac{13}{2}+\frac{7}{2}=-\frac{6}{2}=-3\)

=> x = \(\frac{5}{2}\) or x = \(-\frac{3}{4}\)

Hence, \(\frac{5}{2}\) and \(-\frac{3}{4}\) are the roots of the given equation.

QUESTION-8:

\(7x^{2}+3x-4=0\)

Solution:

\(7x^{2}+3x-4=0\)

\(\Rightarrow 49x^{2}+21x-28=0\) (Multiplying both sides by 7)

\(\Rightarrow 49x^{2}+21x=28\)

\(\Rightarrow \left ( 7x \right )^{2}+2\times 7x\times \frac{3}{2}+\left ( \frac{3}{2} \right )^{2}=28+\left ( \frac{3}{2} \right )^{2}\) [Adding \(\left ( \frac{3}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left ( 7x+\frac{3}{2} \right )^{2}=28+\frac{9}{4}=\frac{121}{4}=\left ( \frac{11}{2} \right )^{2}\)

\(\Rightarrow 7x+\frac{3}{2}=\pm \frac{11}{2}\) (Taking square root on both sides)

\(\Rightarrow 7x+\frac{3}{2}=\frac{11}{2}\:or\:7x+\frac{3}{2}=-\frac{11}{2}\)

\(\Rightarrow 7x=\frac{11}{2}-\frac{3}{2}=\frac{8}{2}=4\:or\:7x=-\frac{11}{2}-\frac{3}{2}=-\frac{14}{2}=-7\)

=> x = \(\frac{4}{7}\) or x = -1

Hence, \(\frac{4}{7}\) and -1 are the roots of the given equation.

QUESTION-9:

\(3x^{2}-2x-1=0\)

Solution:

\(3x^{2}-2x-1=0\)

\(\Rightarrow 9x^{2}-6x-3=0\) (Multiplying both sides by 3)

\(\Rightarrow 9x^{2}-6x=3\)

\(\Rightarrow \left ( 3x \right )^{2}-2\times 1+1^{2}=3+1^{2}\) [Adding \(1^{2}\) on both sides]

\(\Rightarrow \left (3x-1 \right)^{2}=3+1=4=\left ( 2 \right )^{2}\)

\(\Rightarrow 3x-1 =\pm 2\) (Taking square root on both sides)

=> 3x – 1 = 2 or 3x – 1 = -2

=> 3x = 3 or 3x = -1

=> x = 1 or x = \(-\frac{1}{3}\)

Hence, 1 and \(-\frac{1}{3}\) are the roots of the given equation.

QUESTION-10:

\(5x^{2}-6x-2=0\)

Solution:

\(5x^{2}-6x-2=0\)

\(\Rightarrow 25x^{2}-30x-10=0\) (Multiplying both sides by 5)

\(\Rightarrow 25x^{2}-30x=10\)

\(\Rightarrow \left ( 5x \right )^{2}-2\times 5x\times3+3^{2}=10+3^{2}\) [Adding \(3^{2}\) on both sides]

\(\Rightarrow \left ( 5x-3 \right )^{2}=10+9=19\)

\(\Rightarrow 5x-3=\pm \sqrt{19}\) (Taking square root on both sides)

\(\Rightarrow 5x-3= \sqrt{19}\:or\:5x-3=-\sqrt{19}\)

\(\Rightarrow 5x=3+\sqrt{19}\:or\:5x=3-\sqrt{19}\)

\(\Rightarrow x=\frac{3+\sqrt{19}}{5}\:or\:x=\frac{3-\sqrt{19}}{5}\)

Hence, \(\frac{3+\sqrt{19}}{5}\:and\:\frac{3-\sqrt{19}}{5}\) are the roots of the given equation.

QUESTION-11:

\(\frac{2}{x^{2}}-\frac{5}{x}+2=0\)

Solution:

\(\Rightarrow \frac{2-5x+2x^{2}}{x^{2}}=0\)

\(\Rightarrow 2x^{2}-5x+2=0\)

\(\Rightarrow 4x^{2}-10x+4=0\) (Multiplying both sides by 2)

\(\Rightarrow 4x^{2}-10x=-4\)

\(\Rightarrow \left ( 2x \right )^{2}-2\times 2x\times \frac{5}{2}+\left ( \frac{5}{2} \right )^{2}=-4+\left ( \frac{5}{2} \right )^{2}\) [Adding \(\left ( \frac{5}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left ( 2x-\frac{5}{2} \right )^{2}=-4+\frac{25}{4}=\frac{9}{4}=\left ( \frac{3}{2} \right )^{2}\)

\(\Rightarrow 2x-\frac{5}{2} = \pm \frac{3}{2}\) (Taking square root on both sides)

\(\Rightarrow 2x-\frac{5}{2}=\frac{3}{2}\:or\:2x-\frac{5}{2}=-\frac{3}{2}\)

\(\Rightarrow 2x = \frac{3}{2}+\frac{5}{2}=\frac{8}{2}=4\:or\:2x=-\frac{3}{2}+\frac{5}{2}=\frac{2}{2}=1\)

=> x = 2 or x = \(\frac{1}{2}\)

Hence, 2 and \(\frac{1}{2}\) are the roots of the given equation.

QUESTION-12:

\(4x^{2}+4bx-\left ( a^{2}-b^{2} \right )=0\)

Solution:

\(4x^{2}+4bx-\left ( a^{2}-b^{2} \right )=0\)

\(\Rightarrow 4x^{2}+4bx=a^{2}-b^{2}\)

\(\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times b+b^{2}=a^{2}-b^{2}+b^{2}\) [Adding \(b^{2}\) on both sides]

\(\Rightarrow \left ( 2x+b \right )^{2}=a^{2}\)

\(\Rightarrow 2x+b =\pm a\) (Taking square root on both sides)

=> 2x+b = a or 2x+b = -a

=> 2x = a-b or 2x = -a-b

\(\Rightarrow x=\frac{a-b}{2}\:or\:x= -\frac{a+b}{2}\)

Hence, \(\frac{a-b}{2}\:and\:-\frac{a+b}{2}\) are the roots of the given equation.

Question-13:

\(x^{2}-\left ( \sqrt{2}+1 \right )x+\sqrt{2}=0\)

Solution:

\(x^{2}-\left ( \sqrt{2}+1 \right )x+\sqrt{2}=0\)

\(x^{2}-\left ( \sqrt{2}+1 \right )x+\sqrt{2}=-\sqrt{2} \)

\(\Rightarrow x^{2}-2\times x\times \left ( \frac{\sqrt{2}+1}{2} \right )+\left ( \frac{\sqrt{2}+1}{2} \right )^{2}=-\sqrt{2}+\left ( \frac{\sqrt{2}+1}{2} \right )^{2}\) [Adding \(\left ( \frac{\sqrt{2}+1}{2} \right )^{2}\) on both sides]

\(\Rightarrow \left [ x-\left ( \frac{\sqrt{2}+1}{2}^{2} \right ) \right ]^{2}=\frac{-4\sqrt{2}+2+1+2\sqrt{2}}{4}=\frac{2-2\sqrt{2}+1}{4}=\left ( \frac{\sqrt{2}-1}{4} \right )^{2}\)

\(\Rightarrow x-\left ( \frac{\sqrt{2}+1}{2} \right )=\pm \left ( \frac{\sqrt{2}-1}{2} \right )\) (Taking square root on both sides)

\(\Rightarrow x-\left ( \frac{\sqrt{2}+1}{2} \right )=\left ( \frac{\sqrt{2}-1}{2} \right )\:or\:x-\left ( \frac{\sqrt{2}+1}{2} \right )=-\left ( \frac{\sqrt{2}-1}{2} \right )\)

\(\Rightarrow x=\frac{\sqrt{2}+1}{2}+\frac{\sqrt{2}-1}{2}\:or\:x=\frac{\sqrt{2}+1}{2}-\frac{\sqrt{2}-1}{2}\)

\(\Rightarrow x=\frac{2\sqrt{2}}{2}=\sqrt{2}\:or\:x=\frac{2}{2}=1\)

Hence, \(\sqrt{2}\) and 1 are the roots of the given equation.

Question-14:

\(\sqrt{2}x^{2}-3x-2\sqrt{2}=0\)

Solution:

\(\sqrt{2}x^{2}-3x-2\sqrt{2}=0\)

\(\Rightarrow \sqrt{2}x^{2}-3\sqrt{2}x-4=0\) (Multiplying both sides by \(\sqrt{2}\))

\(\Rightarrow \sqrt{2}x^{2}-3\sqrt{2}x=4\)

\(\Rightarrow \left ( \sqrt{2}x \right )^{2}-2\times \sqrt{2}x\times \frac{3}{2}+\left ( \frac{3}{2} \right )^{2}=4+\left ( \frac{3}{2} \right )^{2}\) (Adding \(\left ( \frac{3}{2} \right )^{2}\) on both sides)

\(\Rightarrow \left ( \sqrt{2}x-\frac{3}{2} \right )^{2}=4+\frac{9}{4}=\frac{25}{4}=\left ( \frac{5}{2} \right )^{2}\)

\(\Rightarrow \sqrt{2}x-\frac{3}{2}=\pm \frac{5}{2}\) (Taking square root on both sides)

\(\Rightarrow \sqrt{2}x-\frac{3}{2}=\frac{5}{2}\:or\:\sqrt{2}x-\frac{3}{2}=-\frac{5}{2}\)

\(\Rightarrow \sqrt{2}x=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4\:or\:\sqrt{2}x=-\frac{5}{2}+\frac{3}{2}=-\frac{2}{2}=-1\)

\(\Rightarrow x=\frac{4}{\sqrt{2}}=2\sqrt{2}\:or\:x=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\)

Hence, \(2\sqrt{2}\:and\:-\frac{\sqrt{2}}{2}\) are the roots of the given equation.

Question-15:

\(\sqrt{3}x^{2}+10x+7\sqrt{3}=0\)

Solution:

\(\sqrt{3}x^{2}+10x+7\sqrt{3}=0\)

\(\Rightarrow \sqrt{3}x^{2}+10\sqrt{3}x+21=0\) (Multiplying both sides by \(\sqrt{3}\))

\(\Rightarrow \sqrt{3}x^{2}+10\sqrt{3}x=-21\)

\(\Rightarrow \left ( \sqrt{3x} \right )^{2}+2\times \sqrt{3}x\times 5+5^{2}=-21+5^{2}\) (Adding \(5^{2}\) on both sides)

=> \(\left ( \sqrt{3}x+5 \right )^{2}=-21+25=4=2^{2}\)

=> \(\sqrt{3}x+5=\pm 2\) (Taking square root on both sides)

=> \(\sqrt{3}x+5=2\:or\:\sqrt{3}x+5=-2\)

=> \(\sqrt{3}x=-3\:or\:\sqrt{3}x=-7\)

=> x = \(-\frac{3}{\sqrt{3}}=-\sqrt{3}\:or\:x=-\frac{7}{\sqrt{3}}=-\frac{7\sqrt{3}}{3}\)

Hence, \(-3\:and\:-\frac{7\sqrt{3}}{3}\) are the roots of the given equation.

Question-16:

By using the method of completing the square, show that the equation \(2x^{2}+x+4=0\) has no real roots.

Solution:

\(2x^{2}+x+4=0\)

=> \(4x^{2}+2x+8=0\) (Multiplying both sides by 2)

=> \(4x^{2}+2x=-8\)

\(\Rightarrow \left ( 2x \right )^{2}+2\times 2x\times \frac{1}{2}+\left ( \frac{1}{2} \right )^{2}=-8+\left ( \frac{1}{2} \right )^{2}\) (Adding \(\left ( \frac{1}{2} \right )^{2}\) on both sides)

\(\Rightarrow \left ( 2x+\frac{1}{2} \right )^{2}=-8+\frac{1}{4}=-\frac{31}{4}< 0\)

But, \(\left ( 2x+\frac{1}{2} \right )^{2}\) cannot be negative for any real value of x. So, there is no real value of x satisfying the given equation.

Hence, the given equation has no real roots.


Practise This Question

If the people of india continue polluting the rivers at the same rate as of now. Then what are the worst things that can happen?